FZU2082

树链剖分后要处理的是边的权值，而不是点的权值，但是只要边权下放到点，就可以了

如图

但是问题是，求图4->5路径的权值之和， 那么就会把点3给算进去

那么就要减去，

或者干脆不加进去

有两种方法

 #include <stdio.h>

 #include <string.h>

 #include <stdlib.h>

 #include <algorithm>

 #include <iostream>

 #include <queue>

 #include <stack>

 #include <vector>

 #include <map>

 #include <set>

 #include <string>

 #include <math.h>

 using namespace std;

 #pragma warning(disable:4996)

 #pragma comment(linker, "/STACK:1024000000,1024000000")

 typedef long long LL;

 const int INF = <<;

 /*

 12 1

 1 2 1

 2 3 1

 3 4 1

 4 5 1

 5 11 1

 11 12 1

 1 6 1

 6 7 1

 7 8 1

 6 9 1

 6 10 1

 1 9 10

 */

 const int N =  + ;

 vector<int> g[N];

 int size[N], son[N], fa[N], depth[N], top[N], w[N], num, ra[N];

 LL tree[N*];

 int edge[N][];

 int h[N];

 void dfs(int u)

 {

     size[u] = ;

     son[u] = ;

     for (int i = ; i < g[u].size(); ++i)

     {

         int v = g[u][i];

         if (v != fa[u])

         {

             fa[v] = u;

             depth[v] = depth[u] + ;

             dfs(v);

             size[u] += size[v];

             if (size[v] > size[son[u]])

                 son[u] = v;

         }

     }

 }

 void dfs2(int u, int tp)

 {

     w[u] = ++num;

     ra[num] = u;

     top[u] = tp;

     if (son[u] != )

         dfs2(son[u], top[u]);

     for (int i = ; i < g[u].size(); ++i)

     {

         int v = g[u][i];

         if (v != fa[u] && v != son[u])

             dfs2(v, v);

     }

 }

 void pushUp(int rt)

 {

     tree[rt] = tree[rt << ] + tree[rt <<  | ];

 }

 void update(int l, int r, int rt, int pos, int val)

 {

     if (l == r)

     {

         tree[rt] = val;

         return;

     }

     int mid = (l + r) >> ;

     if (pos <= mid)

         update(l, mid, rt << , pos, val);

     else

         update(mid + , r, rt <<  | , pos, val);

     pushUp(rt);

 }

 LL ans;

 void query(int l, int r, int rt, int L, int R)

 {

     if (L <= l&&R >= r)

     {

         ans += tree[rt];

         return;

     }

     int mid = (l + r) >> ;

     if (L <= mid)

         query(l, mid, rt << , L, R);

     if (R > mid)

         query(mid + , r, rt <<  | , L, R);

 }

 int main()

 {

     int n, m, a, b, c;

     while (scanf("%d%d", &n, &m) != EOF)

     {

         memset(tree, , sizeof(tree));

         for (int i = ; i <= n; ++i)

             g[i].clear();

         num = ;

         for (int i = ; i < n; ++i)

         {

             scanf("%d%d%d", &edge[i][], &edge[i][], &edge[i][]);

             g[edge[i][]].push_back(edge[i][]);

             g[edge[i][]].push_back(edge[i][]);

         }

         fa[] = depth[] = ;

         dfs();

         dfs2(, );

         for (int i = ; i < n; ++i)

         {

             if (depth[edge[i][]] > depth[edge[i][]])

                 swap(edge[i][], edge[i][]);

             h[edge[i][]] = edge[i][];

             update(, n, , w[edge[i][]], edge[i][]);

         }

         while (m--)

         {

             scanf("%d%d%d", &a, &b, &c);

             if (a == )

             {

                 if (depth[edge[b][]] > depth[edge[b][]])

                     swap(edge[b][], edge[b][]);

                 update(, n, , w[edge[b][]], c);

                 h[edge[b][]] = c;

             }

             else

             {

                 ans = ;

                 while (top[b] != top[c])

                 {

                     if (depth[top[b]] < depth[top[c]])

                         swap(b, c);

                     query(, n, , w[top[b]], w[b]);

                     b = fa[top[b]];

                 }

                 //b和c一定是在一条链上

                 if (depth[b]>depth[c])

                     swap(b, c);

                 //如果b有父亲，那么就有边下放到它，多加了这个点，要减去

                 query(, n, , w[b], w[c]);

                 if (fa[b] != )

                     ans -= h[b];

                 /*

                 如果b，c是在一条重链上， 那么询问w[son[b]] 到w[c]

                 那么就没有加到点b

                 否则，b，c就是指向同一点，那么w[son[b]] > w[c] , 询问是不会增加任何值的

                 query(1, n, 1, w[son[b]], w[c]);

                 */

                 printf("%lld\n", ans);

             }

         }

     }

     return ;

 }